# Cross product

The cross product is an operation that takes two nonzero vectors and produces a *vector* (not a scalar) perpendicular to both of them. Geometrically, we define the magnitude of the cross product with

$|\overrightarrow{u}\times \overrightarrow{v}|=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta \text{,}$

where $\theta $ is the angle between the vectors in standard position. To find the direction of the vector, we use the *right-hand rule*: point in the direction of $\overrightarrow{u}$ with your fingers, and then curl them (naturally, not backwards) towards the direction of $\overrightarrow{v}\text{.}$ Your thumb will then point in the direction of $\overrightarrow{u}\times \overrightarrow{v}\text{.}$

Algebraically, we define the cross product in ${\mathbb{R}}^{3}$ with

$[{u}_{1},{u}_{2},{u}_{3}]\times [{v}_{1},{v}_{2},{v}_{3}]=[{w}_{1},{w}_{2},{w}_{3}]\text{,}$

where the components are given by

Component | Value | Remember |
---|---|---|

${w}_{1}$ | ${u}_{2}{v}_{3}-{u}_{3}{v}_{2}$ | 2, 3 |

${w}_{2}$ | ${u}_{3}{v}_{1}-{u}_{1}{v}_{3}$ | 3, 1 |

${w}_{3}$ | ${u}_{1}{v}_{2}-{u}_{2}{v}_{1}$ | 1, 2 |

The pattern shouldn’t be too hard to see. Each part has a product of two terms, and then you just subtract the same two terms with the subscripts swapped. You essentially need to remember the sequence 233112, and even that has some repetition. You do not need to remember anything for ${\mathbb{R}}^{2}\text{,}$ since the cross product exists only in ${\mathbb{R}}^{3}$ and in ${\mathbb{R}}^{7}\text{.}$

Like the dot product, the cross product has some important properties. First, it can produce the zero vector:

$\overrightarrow{v}\times \overrightarrow{0}=\overrightarrow{0}$ and $\overrightarrow{v}\times \overrightarrow{v}=\overrightarrow{0}\text{.}$

Unlike the dot product, is is anticommutative—order matters:

$\overrightarrow{u}\times \overrightarrow{v}=-(\overrightarrow{u}\times \overrightarrow{v})\text{.}$

It is distributive over addition:

$\overrightarrow{u}\times (\overrightarrow{v}+\overrightarrow{w})=\overrightarrow{u}\times \overrightarrow{v}+\overrightarrow{u}\times \overrightarrow{w}\text{.}$

It is associative over scalar multiplication:

$k\overrightarrow{u}\times \overrightarrow{v}=k(\overrightarrow{u}\times \overrightarrow{v})=\overrightarrow{u}\times k\overrightarrow{v}\text{.}$

We can use the dot product and cross product together to make a test for coplanarity. Here it is: vectors $\overrightarrow{u}\text{,}$ $\overrightarrow{v}\text{,}$ and $\overrightarrow{w}$ are coplanar if and only if

$\overrightarrow{u}\cdot \overrightarrow{v}\times \overrightarrow{w}=0\text{.}$

It does not matter which vectors are placed where in that equation. As long as crossing two of the vectors and dotting the result with the third yields zero, they are coplanar. Why does this work? When we cross $\overrightarrow{v}$ and $\overrightarrow{w}\text{,}$ we get a vector perpendicular to both of them. If $\overrightarrow{u}$ is on the same plane, it should *also* be perpendicular to the cross product. We test for this using the dot product—two vectors are perpendicular if their dot product is zero.